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这段代码有问题吗？我不断收到编译错误。基本上我想将一个 void 返回函数连接到一个具有非 void 返回类型的信号。增强版：1.46.1 版

#include <boost/signals2.hpp>
#include <boost/lambda/bind.hpp>
#include <boost/lambda/lambda.hpp>
using namespace boost::signals2;

void func()
{
  printf("Func called!");
}

main()
{
  signal<int(int)> sig;
  sig.connect( (boost::lambda::bind(func),  1) );
}

编译时出现以下错误：

/opt/include/boost/signals2/detail/slot_template.hpp: In member function ‘void boost::signals2::slot1<R, T1, SlotFunction>::init_slot_function(const F&) [with F = int, R = int, T1 = int, SlotFunction = boost::function<int(int)>]’:
/opt/include/boost/signals2/detail/slot_template.hpp:81:9:   instantiated from ‘boost::signals2::slot1<R, T1, SlotFunction>::slot1(const F&) [with F = int, R = int, T1 = int, SlotFunction = boost::function<int(int)>]’
hello-world-example.cpp:13:51:   instantiated from here
/opt/include/boost/signals2/detail/slot_template.hpp:156:9: error: invalid conversion from ‘int’ to ‘boost::function<int(int)>::clear_type*’ [-fpermissive]
/opt/include/boost/function/function_template.hpp:1110:14: error:   initializing argument 1 of ‘boost::function<R(T0)>::self_type& boost::function<R(T0)>::operator=(boost::function<R(T0)>::clear_type*) [with R = int, T0 = int, boost::function<R(T0)>::self_type = boost::function<int(int)>]’ [-fpermissive]

谢谢。

4

1 回答 1

0

似乎在 Boost 版本 1.49 上编译没有问题

于 2012-08-17T10:09:04.170 回答